Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x - 7} = \dfrac{-10x - 30}{x - 7}$
Explanation: Multiply both sides by $x - 7$ $ \dfrac{x^2 + x}{x - 7} (x - 7) = \dfrac{-10x - 30}{x - 7} (x - 7)$ $ x^2 + x = -10x - 30$ Subtract $-10x - 30$ from both sides: $ x^2 + x - (-10x - 30) = -10x - 30 - (-10x - 30)$ $ x^2 + x + 10x + 30 = 0$ $ x^2 + 11x + 30 = 0$ Factor the expression: $ (x + 6)(x + 5) = 0$ Therefore $x = -6$ or $x = -5$ The original expression is defined at $x = -6$ and $x = -5$, so there are no extraneous solutions.